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Determinants

Question
CBSEENMA12034708
Wired Faculty App

Examine the consistency of the system of equations:
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3

Solution

The given equations are
3x – y – 2z = 2
0x + 2y – z = – 1
3x – 5y+Oz = 3
These equations can be written as
open square brackets table row 3 cell space space space minus 1 end cell cell space space space minus 2 end cell row 0 cell space space space space space space 2 end cell cell space space space minus 1 end cell row 3 cell space space minus 5 end cell cell space space space space space space 0 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell space space space 2 end cell row cell negative 1 end cell row cell space space 3 end cell end table close square brackets
therefore space space space space space space AX space equals space straight B space space where space straight A space equals space open square brackets table row 3 cell space minus 1 end cell cell space space minus 2 end cell row 0 cell space space space space space 2 end cell cell space space minus 1 end cell row 3 cell space minus 5 end cell cell space space space 0 end cell end table close square brackets comma space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space straight B space equals space open square brackets table row cell space space space 2 end cell row cell negative 1 end cell row cell space space space 3 end cell end table close square brackets space space space space space space space space space space space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space minus 1 end cell cell space space minus 2 end cell row 0 cell space space space 2 end cell cell space space minus 1 end cell row 3 cell space space minus 5 end cell cell space space space space space 0 end cell end table close vertical bar space equals space 3 open vertical bar table row 2 cell space space minus 1 end cell row cell negative 5 end cell cell space space space space space 0 end cell end table close vertical bar space minus 0 space open vertical bar table row cell negative 1 end cell cell space space minus 2 end cell row cell negative 5 end cell cell space space space space 0 end cell end table close vertical bar plus 3 space open vertical bar table row cell negative 1 end cell cell space space minus 2 end cell row 2 cell space minus 1 end cell end table close vertical bar space space space space space space space space space space space space space space space space space space space space space equals 3 left parenthesis 0 minus 5 right parenthesis minus 0 plus 3 left parenthesis 1 plus 4 right parenthesis space equals space minus 15 minus 0 plus 15 space equals space 0 therefore space space space space space straight A to the power of negative 1 end exponent space does space not space exists. space space space space space
Cofactors of the elements of first row of | A | are
open vertical bar table row cell space space 2 end cell cell space space space minus 1 end cell row cell negative 5 end cell cell space space space space space space 0 end cell end table close vertical bar comma space space space minus open vertical bar table row 0 cell space space space minus 1 end cell row 3 cell space space space space space 0 end cell end table close vertical bar comma space space open vertical bar table row 0 cell space space space space 2 end cell row 3 cell space space minus 5 end cell end table close vertical bar
i.e. 0 – 5, – (0 + 3), 0 – 6 i.e. – 5, – 3, – 6 respectively.
Cofactors of the elements of second row of | A | are
negative open vertical bar table row cell negative 1 end cell cell space space space minus 2 end cell row cell negative 5 end cell cell space space space space 0 end cell end table close vertical bar comma space space space space open vertical bar table row 3 cell space space space minus 2 end cell row 3 cell space space space space space 0 end cell end table close vertical bar comma space space minus open vertical bar table row 3 cell space space space minus 1 end cell row 3 cell space space minus 5 end cell end table close vertical bar
i.e. – (0 – 10), 0 + 6, –(– 15 + 3) i.e. 10, 6, 12 respectively.
Co-factors of the elements of third row of | A | are
open vertical bar table row cell negative 1 end cell cell space space minus 2 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row 3 cell space space minus 2 end cell row 0 cell space space minus 1 end cell end table close vertical bar comma space space open vertical bar table row 3 cell space space minus 1 end cell row 0 cell space space space space 2 end cell end table close vertical bar
i.e. 1 + 4, – (– 3 – 0), 6 – 0 i.e. 5, 3, 6 respectively.
therefore space space adj space straight A space equals space open square brackets table row cell negative 5 end cell cell space space minus 3 end cell cell space space minus 6 end cell row 10 cell space space space space 6 end cell cell space space space 12 end cell row 5 cell space space 3 end cell cell space space 6 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 5 end cell cell space space 10 end cell cell space space space space 5 end cell row cell negative 3 end cell cell space space 6 end cell cell space space space 3 end cell row cell negative 6 end cell cell space space 12 end cell cell space space space 6 end cell end table close square brackets therefore space space space left parenthesis adj space straight A right parenthesis thin space straight B space equals open square brackets table row cell negative 5 end cell cell space space 10 end cell cell space space 5 end cell row 10 6 cell space space 3 end cell row 5 12 cell space space 6 end cell end table close square brackets space open square brackets table row cell space space space 2 end cell row cell negative 1 end cell row cell space space space 3 end cell end table close square brackets space equals space open square brackets table row cell negative 10 minus 10 plus 15 end cell row cell negative 6 minus 6 plus 9 end cell row cell negative 12 minus 12 plus 18 end cell end table close square brackets space equals space open square brackets table row cell negative 5 end cell row cell negative 3 end cell row 6 end table close square brackets space not equal to space straight O
∴ solution does not exist and so system of equations is inconsistent.

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